Joe’s pie shop serves only chocolate pies and coconut cream pies. On a

Please refer the affixed diagram.

St1:-Customers who didn’t buy chocolate pie=customers who bought only coconut pie=b=40
Customers who bought chocolate pie=a+c=200-40=160
So, Joe sold 160 nos. of chocolate pies.
Sufficient.
St2:- 120 customers bought coconut pie, b + c=120 or, b=120-80=40
So, a+c=200-b=200-40=160.
So, Joe sold 160 nos. of chocolate pies.
Sufficient.

Ans. (D)

Davidtutor's explanation is correct, even if not very detailed. The other explanations saying that the result is 160 are wrong.
They confuse pies and clients.
200 are pies, but the rest of the numbers refer to clients.
So if 80 clients bought both, it means that we have 80 choc pies and 80 coco pies.
200-2x80= 40 remaining pies.

St1) 40 didnt buy choc, so bought only coco, so the 40 remaining pies are all coco, and the choc are 80
st2) 120 bought a coco. Minus the 80 who bought both, we have 40 that bought only coco, and the choc are 80.

Answer D.

Hi gioacchinorossini,
Please go through Davidtutor's explanation again, it says 160 #customers bought chocolate pies.

Dear PKN, I dont see where Davidtutor says "160 #customers bought chocolate pies" (that would be wrong).
His explanation is correct in my opinion. The other explanations of the other users are wrong (even if they come to the same answer D).

[/quote]
Dear PKN, I dont see where Davidtutor says "160 #customers bought chocolate pies" (that would be wrong).
His explanation is correct in my opinion. The other explanations of the other users are wrong (even if they come to the same answer D).[/quote]

Hi gioacchinorossini
Extended explanation:-

2 overlapping sets can always be broken into the same 4 categories (both, only A, only B, neither).
This is a Precise approach.

total number of pies= 200
(both chocolate and cocount) = 80
neither = 0 as all customers bought at least one pie
so (only chocolate) + (only cococunt) = total - (both) - (neither) = 200 - 80 - 0 = 120
question - how much is (only chocolate) + (both chocolate and coconut)?

Since we know that (only chocolate) + (only coconut) = 120,----(1) all we need to is the value of (only coconut).

(1) then these 40 must have bought a cococunt pie meaning that (only coconut) = 40. Exactly what we need!

Now, from eq(1), Only chocoloate=120-40=80
Therefore, (only chocolate) + (both chocolate and coconut)=80+80=160
Sufficient.

(2) gives us (only coconut) + (both). Since we know that (both) = 80, we can calculate (only coconut).
So, Only Coconut=120-80=40
Now, from eq(1), Only chocoloate=120-40=80
Therefore, (only chocolate) + (both chocolate and coconut)=80+80=160
Sufficient.

Note:- In DS questions, exact computation is not required. Mr. David did the same thru reasoning, final answer(numerical value) is not required , only data sufficiency is checked.

I disagree. There are no choco-coco pies. The pies are either choco or coco. But the clients can buy coco or choco or both. That's why I said that all explanations, apart from the David's one, are wrong, because they confuse clients and pies, and they add clients and pies. I know that in DS you dont need to get the result, I am a GMAT tutor since 2009. Still, if we want to find the answer as if it were PS, it is 80 and not 160.

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